Editing Maximum power transfer theorem (section)

=== Proof ===

[[File:Source and load boxes.png|left|source and load impedance diagram]]

In this diagram, [[AC power]] is being transferred from the source, with [[phasor (electronics)|phasor]] magnitude of voltage <math>|V_\text{S}|</math> (positive peak voltage) and fixed [[source impedance]] <math>Z_\text{S}</math> (S for source), to a load with impedance <math>Z_\text{L}</math> (L for load), resulting in a (positive) magnitude <math>|I|</math> of the current phasor <math>I</math>.  This magnitude <math>|I|</math> results from dividing the magnitude of the source voltage by the magnitude of the total circuit impedance:
<math display="block">|I| =  {|V_\text{S}| \over |Z_\text{S} + Z_\text{L}|} .</math>

The average power <math>P_\text{L}</math> dissipated in the load is the square of the current multiplied by the resistive portion (the real part) <math>R_\text{L}</math> of the load impedance <math>Z_\text{L}</math>:
<math display="block">\begin{align}
P_\text{L} & = I_\text{rms}^2 R_\text{L} = {1 \over 2} |I|^2 R_\text{L}\\
& = {1 \over 2} \left( {|V_\text{S}| \over |Z_\text{S} + Z_\text{L}|} \right)^2 R_\text{L}  = {1 \over 2}{ |V_\text{S}|^2 R_\text{L} \over (R_\text{S} + R_\text{L})^2 + (X_\text{S} + X_\text{L})^2},
\end{align}</math>
where <math>R_\text{S}</math> and <math>R_\text{L}</math> denote the resistances, that is the real parts, and  <math>X_\text{S}</math> and <math>X_\text{L}</math> denote the reactances, that is the imaginary parts, of respectively the source and load impedances <math>Z_\text{S}</math> and <math>Z_\text{L}</math>.

To determine, for a given source, the voltage <math>V_\text{S}</math> and the impedance <math>Z_\text{S},</math> the value of the load impedance <math>Z_\text{L}, </math> for which this expression for the power yields a maximum, one first finds, for each fixed positive value of <math>R_\text{L}</math>, the value of the reactive term <math>X_\text{L}</math> for which the denominator:
<math display="block">(R_\text{S} + R_\text{L})^2 + (X_\text{S} + X_\text{L})^2 </math>
is a minimum. Since reactances can be negative, this is achieved by adapting the load reactance to:
<math display="block">X_\text{L} = -X_\text{S}.</math>

This reduces the above equation to:
<math display="block">P_\text{L} = \frac 1 2 \frac{|V_\text{S}|^2 R_\text{L}}{(R_\text{S} + R_\text{L})^2}</math>
and it remains to find the value of <math>R_\text{L}</math> which maximizes this expression. This problem has the same form as in the purely resistive case, and the maximizing condition therefore is <math>R_\text{L} = R_\text{S}.</math>

The two maximizing conditions:

*<math>R_\text{L} =  R_\text{S}</math>
*<math>X_\text{L} = -X_\text{S}</math>

describe the [[complex conjugate]] of the source impedance, denoted by <math>^*,</math> and thus can be concisely combined to:
<math display="block">Z_\text{L} = Z_\text{S}^*.</math>